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2 years ago

In a volleyball match Hanein serves the volleyball The height of the the ball above the court

Mathematics

1 answer:

forsale [732]2 years ago

7 0

(a) The maximum height of the ball above the ground is 12.5 m and the time of motion is 1.43 s.

(b) The time taken for the ball to contact the other player at 0.5 m above the ground is 3.0 s.

<h3>Maximum height reached by the ball</h3>

The maximum height reached by the ball is calculated as follows.

At maximum height, the final velocity, v = 0

dh/dt = v = 0

dh/dt = -2(4.9)t + 14

0 = -9.8t + 14

9.8t = 14

t = 1.43 s

H(1.43) = -4.9(1.43)² + 14(1.43) + 2.5

H(1.43) = -10.02 + 20.02 + 2.5

H(1.43) = 12.5 m

<h3>Time to reach maximum height</h3>

12.5 = -4.9t² + 14t + 2.5

4.9t² - 14t + 10 = 0

t = 1.43 s

<h3>Time for the ball to reach 0.5 m above the ground</h3>

0.5 = -4.9t² + 14t + 2.5

4.9t² - 14t + - 2 = 0

t = 3.0 seconds

The complete question is below:

In a volleyball match Hanein serves the volleyball at 14 m/s, from a height of 2.5 m above the court. The height of the ball in flight is estimated using the equation, h = -4.9t² + 14t + 2.5, where t is the time in second and h is height above ground, in metres.

Learn more about time of motion here: brainly.com/question/2364404

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3 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$

$Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)$

$(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}$

$Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}$

$Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}$

$n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2$

$n = (16*(\frac{1.2}{(270-264)}))^2$

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

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What is the sum of the exterior angles of a convex polygon?

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Answer:

Step-by-step explanation:

The sum of the exterior angles for each polygon is always 360°.

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If you have n-sided convex polygon, then

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